node.js 在 Express 中对所有路径使用特定的中间件，但特定路径除外
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Use specific middleware in Express for all paths except a specific one
提问by Thomas
I am using the Express framework in node.js with some middleware functions:
我在 node.js 中使用 Express 框架和一些中间件功能：
var app = express.createServer(options);
app.use(User.checkUser);
I can use the .usefunction with an additional parameter to use this middleware only on specific paths:
我可以使用.use带有附加参数的函数来仅在特定路径上使用此中间件：
app.use('/userdata', User.checkUser);
Is it possible to use the path variable so that the middleware is used for all paths except a specific one, i.e. the root path?
是否可以使用路径变量，以便中间件用于除特定路径之外的所有路径，即根路径？
I am thinking about something like this:
我在想这样的事情：
app.use('!/', User.checkUser);
So User.checkUseris always called except for the root path.
SoUser.checkUser总是被调用，除了根路径。
回答by chovy
I would add checkUser middleware to all my paths, except homepage.
我会将 checkUser 中间件添加到我的所有路径中，除了主页。
app.get('/', routes.index);
app.get('/account', checkUser, routes.account);
or
或者
app.all('*', checkUser);
function checkUser(req, res, next) {
if ( req.path == '/') return next();
//authenticate user
next();
}
You could extend this with underscore to search for the req.path in an array of non-authenticated paths:
您可以使用下划线对其进行扩展，以在未经身份验证的路径数组中搜索 req.path：
function checkUser(req, res, next) {
var _ = require('underscore')
, nonSecurePaths = ['/', '/about', '/contact'];
if ( _.contains(nonSecurePaths, req.path) ) return next();
//authenticate user
next();
}
回答by jsalonen
Instead of directly registering User.checkUseras middleware, register a new helper function, say checkUserFilter, that gets called on every URL, but passed execution to userFiled` only on given URLs. Example:
不是直接注册User.checkUser为中间件，而是注册一个新的辅助函数，例如checkUserFilter，在每个 URL 上调用，但仅在给定的 URL 上将执行传递给 userFiled`。例子：
var checkUserFilter = function(req, res, next) {
if(req._parsedUrl.pathname === '/') {
next();
} else {
User.checkUser(req, res, next);
}
}
app.use(checkUserFilter);
In theory, you could provide regexp paths to app.use. For instance something like:
理论上，您可以提供到app.use. 例如类似的东西：
app.use(/^\/.+$/, checkUser);
Tried it on express 3.0.0rc5, but it doesn't work.
在 express 3.0.0rc5 上尝试过，但不起作用。
Maybe we could open a new ticketand suggest this as a feature?
也许我们可以开一张新票并建议将此作为一项功能？
回答by Jonatan Lundqvist Medén
You can set the middleware on each route also.
您也可以在每个路由上设置中间件。
// create application/x-www-form-urlencoded parser
var urlencodedParser = bodyParser.urlencoded({ extended: false })
// POST /login gets urlencoded bodies
app.post('/login', urlencodedParser, function (req, res) {
if (!req.body) return res.sendStatus(400)
res.send('welcome, ' + req.body.username)
})
回答by user1872904
Use
用
app.use(/^(\/.+|(?!\/).*)$/, function(req, resp, next){...
This pass any url apart from /. Unless, it works for me.
这会传递除 / 之外的任何 url。除非，它对我有用。
In general
一般来说
/^(\/path.+|(?!\/path).*)$/
(see How to negate specific word in regex?)
（请参阅如何否定正则表达式中的特定单词？）
Hope this helps
希望这可以帮助
回答by Abhay Shiro
Use this library called express-unless
使用这个名为express-unless 的库
Require authentication for every request unless the path is index.html.
除非路径是 index.html，否则要求对每个请求进行身份验证。
app.use(requiresAuth.unless({
path: [
'/index.html',
{ url: '/', methods: ['GET', 'PUT'] }
]
}))
Path it could be a string, a regexp or an array of any of those. It also could be an array of object which is URL and methods key-pairs. If the request path or path and method match, the middleware will not run.
Path 它可以是字符串、正则表达式或其中任何一个的数组。它也可以是一个对象数组，它是 URL 和方法密钥对。如果请求路径或路径与方法匹配，则中间件将不会运行。
This library will surely help you.
这个图书馆肯定会帮助你。
